Copyright © 2011 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Chapter 16 Relational Database Design Algorithms and Further Dependencies

Copyright © 2011 Ramez Elmasri and Shamkant Navathe Inference Rules for Functional Dependencies Definition. Formally, the set of all dependencies that include F as well as all dependencies that can be inferred from F is called the closure of F; it is denoted by F +. An FD X → Y is inferred from a set of dependencies F specified on R if X → Y holds in every legal relation state r of R; that is, whenever r satisfies all the dependencies in F, X → Y also holds in r. The closure F + of F is the set of all functional dependencies that can be inferred from F.

Copyright © 2011 Ramez Elmasri and Shamkant Navathe Inference Rules for Functional Dependencies IR1 (reflexive rule)1: If X ⊇ Y, then X →Y. a set of attributes always determines itself or any of its subsets Proof of IR1. Suppose that X ⊇ Y and that two tuples t1 and t2 exist in some relation instance r of R such that t1 [X] = t2 [X]. Then t1[Y] = t2[Y] because X ⊇ Y; hence, X → Y must hold in r. IR2 (augmentation rule)2: {X → Y} |=XZ → YZ. adding the same set of attributes to both the left- and right- hand sides of a dependency results in another valid dependency. Proof of IR2 (by contradiction). Assume that X → Y holds in a relation instance r of R but that XZ → YZ does not hold. Then there must exist two tuples t1 and t2 in r such that (1) t1 [X] = t2 [X], (2) t1 [Y] = t2 [Y], (3) t1 [XZ] = t2 [XZ], and (4) t1 [YZ] ≠ t2 [YZ]. This is not possible because from (1) and (3) we deduce (5) t1 [Z] = t2 [Z], and from (2) and (5) we deduce (6) t1 [YZ] = t2 [YZ], contradicting (4).

Copyright © 2011 Ramez Elmasri and Shamkant Navathe Inference Rules for Functional Dependencies IR3 (transitive rule): {X → Y, Y → Z} |=X → Z. functional dependencies are transitive Proof of I R3. Assume that (1) X → Y and (2) Y → Z both hold in a relation r. Then for any two tuples t1 and t2 in r such that t1 [X] = t2 [X], we must have (3) t1 [Y] = t2 [Y], from assumption (1); hence we must also have (4) t1 [Z] = t2 [Z] from (3) and assumption (2); thus X → Z must hold in r. IR4 (decomposition, or projective, rule): {X → YZ} |=X → Y. we can remove attributes from the right-hand side of a dependency. decompose the FD X → {A1, A2,..., An} into the set of dependencies {X → A1, X → A2,...,X → An}. Proof of IR4 (Using I R1 through I R3). 1. X → YZ(given). 2. YZ → Y(using IR1 and knowing that YZ ⊇ Y). 3. X → Y(using IR3 on 1 and 2).

Copyright © 2011 Ramez Elmasri and Shamkant Navathe Inference Rules for Functional Dependencies IR5 (union, or additive, rule): {X → Y, X → Z} |=X → YZ. combine a set of dependencies {X → A1, X → A2,..., X → An} into the single FD X → {A1, A2,..., An}. Proof of I R5 (using I R1 through I R3). 1. X →Y(given). 2. X → Z(given). 3. X → XY(using IR2 on 1 by augmenting with X; notice that XX= X). 4. XY → YZ(using IR2 on 2 by augmenting with Y). 5. X → YZ(using IR3 on 3 and 4). IR6 (pseudotransitive rule): {X → Y, WY → Z} |=WX → Z. allows us to replace a set of attributes Yon the left hand side of a dependency with another set X that functionally determines Y Proof of I R6 (using I R1 through I R3). 1. X → Y(given). 2. WY → Z(given). 3. WX → WY(using IR2 on 1 by augmenting with W). 4. WX → Z(using IR3 on 3 and 2).

Copyright © 2011 Ramez Elmasri and Shamkant Navathe Inference Rules for Functional Dependencies Definition. For each such set of attributes X, we determine the set X + of attributes that are functionally determined by X based on F; X + is called the closure of X under F. Algorithm Determining X + the Closure of X under F Input: A set F of FDs on a relation schema R, and a set of attributes X, which is a subset of R. X + := X; repeat oldX + := X + ; for each functional dependency Y → Z in F do If X + ⊇ Y then X + := X + ∪ Z; until (X + = oldX + );

Copyright © 2011 Ramez Elmasri and Shamkant Navathe Inference Rules for Functional Dependencies F = {Ssn → Ename, Pnumber → {Pname, Plocation}, {Ssn, Pnumber} → Hours} Using Algorithm 16.1, we calculate the following closure sets with respect to F: {Ssn} + = {Ssn, Ename} {Pnumber} + = {Pnumber, Pname, Plocation} {Ssn, Pnumber} + = {Ssn, Pnumber, Ename, Pname, Plocation, Hours}

Copyright © 2011 Ramez Elmasri and Shamkant Navathe Equivalence of Sets of Functional Dependencies Definition. A set of functional dependencies F is said to cover another set of functional dependencies E if every FD in E is also in F + ; that is, if every dependency in E can be inferred from F; alternatively, we can say that E is covered by F. Definition. Two sets of functional dependencies E and F are equivalent if E + = F +. Therefore, equivalence means that every FD in E can be inferred from F, and every FD in F can be inferred from E; that is, E is equivalent to F if both the conditions—E covers F and F covers E—hold. We can determine whether F covers E by calculating X + with respect to F for each FD X → Y in E, and then checking whether this X + includes the attributes in Y. If this is the case for every FD in E, then F covers E. We determine whether E and F are equivalent by checking that E covers F and F covers E

Copyright © 2011 Ramez Elmasri and Shamkant Navathe Minimal Sets of Functional Dependencies A set of functional dependencies F to be minimal if it satisfies the following conditions: 1. Every dependency in F has a single attribute for its right- hand side. 2. We cannot replace any dependency X → A in F with a dependency Y → A, where Y is a proper subset of X, and still have a set of dependencies that is equivalent to F. 3. We cannot remove any dependency from F and still have a set of dependencies that is equivalent to F. Definition. A minimal cover of a set of functional dependencies E is a minimal set of dependencies (in the standard canonical form and without redundancy) that is equivalent to E

Copyright © 2011 Ramez Elmasri and Shamkant Navathe Minimal Sets of Functional Dependencies Algorithm Finding a Minimal Cover F for a Set of Functional Dependencies E Input: A set of functional dependencies E. 1. Set F := E. 2. Replace each functional dependency X → {A1, A2,..., An} in F by the n functional dependencies X →A1, X →A2,..., X → An. 3. For each functional dependency X → A in F for each attribute B that is an element of X if { {F – {X → A} } ∪ { (X– {B} ) → A} } is equivalent to F then replace X → A with (X– {B} ) → A in F. 4. For each remaining functional dependency X → A in F if{F – {X → A} } is equivalent to F, then remove X → A from F.

Copyright © 2011 Ramez Elmasri and Shamkant Navathe Minimal Sets of Functional Dependencies Let the given set of FDs be E : {B → A, D → A, AB → D}. We have to find the minimal cover of E. Step 2 done, as All above dependencies are in canonical form. Since B → A, by augmenting with B on both sides (IR2), we have BB → AB, or B → AB (i). However, AB → D as given (ii). By the transitive rule (I R3), we get from (i) and (ii), B → D. Thus AB → D may be replaced by B → D. We now have a set equivalent to original E, say E’: {B → A, D → A, B → D}. No further reduction is possible in step 3 since all FDs have a single attribute on the left-hand side. In step 4 we look for a redundant FD in E’. By using the transitive rule on B → D and D → A, we derive B → A. Hence B → A is redundant in E’ and can be eliminated. Therefore, the minimal cover of E is {B → D, D → A}.

Copyright © 2011 Ramez Elmasri and Shamkant Navathe Finding a Key K for R Algorithm 16.2(a). Finding a Key K for R Given a set F of Functional Dependencies Input: A relation R and a set of functional dependencies F on the attributes of R. 1. Set K:= R. 2. For each attribute A in K {compute (K– A) + with respect to F; if(K – A) + contains all the attributes in R, then set K := K – {A} };

Copyright © 2011 Ramez Elmasri and Shamkant Navathe Relation Decomposition and Insufficiency of Normal Forms Universal Relation Schema: A relation schema R = {A1, A2, …, An} that includes all the attributes of the database. Universal relation assumption: Every attribute name is unique. Decomposition: The process of decomposing the universal relation schema R into a set of relation schemas D = {R1,R2, …, Rm} that will become the relational database schema by using the functional dependencies. Attribute preservation condition: Each attribute in R will appear in at least one relation schema Ri in the decomposition so that no attributes are “lost”. Another goal of decomposition is to have each individual relation Ri in the decomposition D be in BCNF or 3NF. Additional properties of decomposition are needed to prevent from generating spurious tuples

Copyright © 2011 Ramez Elmasri and Shamkant Navathe Lossless (Non-additive) Join Property of a Decomposition Definition: Lossless join property: a decomposition D = {R1, R2,..., Rm} of R has the lossless (nonadditive) join property with respect to the set of dependencies F on R if, for every relation state r of R that satisfies F, the following holds, where * is the natural join of all the relations in D: * (  R1 (r),...,  Rm (r)) = r Note: The word loss in lossless refers to loss of information, not to loss of tuples. In fact, for “loss of information” a better term is “addition of spurious information”

Copyright © 2011 Ramez Elmasri and Shamkant Navathe Lossless (Non-additive) Join Property of a Decomposition (cont.) Algorithm 16.3: Testing for Lossless Join Property Input: A universal relation R, a decomposition D = {R1, R2,..., Rm} of R, and a set F of functional dependencies. 1. Create an initial matrix S with one row i for each relation Ri in D, and one column j for each attribute Aj in R. 2. Set S(i,j):=bij for all matrix entries. (* each bij is a distinct symbol associated with indices (i,j) *). 3. For each row i representing relation schema Ri {for each column j representing attribute Aj {if (relation Ri includes attribute Aj) then set S(i,j):= aj;};}; (* each aj is a distinct symbol associated with index (j) *) 4. Repeat the following loop until a complete loop execution results in no changes to S {for each functional dependency X  Y in F {for all rows in S which have the same symbols in the columns corresponding to attributes in X {make the symbols in each column that correspond to an attribute in Y be the same in all these rows as follows: if any of the rows has an “a” symbol for the column, set the other rows to that same “a” symbol in the column. If no “a” symbol exists for the attribute in any of the rows, choose one of the “b” symbols that appear in one of the rows for the attribute and set the other rows to that same “b” symbol in the column ;}; }; 5. If a row is made up entirely of “a” symbols, then the decomposition has the lossless join property; otherwise it does not.

Copyright © 2011 Ramez Elmasri and Shamkant Navathe Lossless (Non-additive) Join Property of a Decomposition (a) Case 1: Decomposition of EMP_PROJ into EMP_PROJ1 and EMP_LOCS fails test. (b) A decomposition of EMP_PROJ that has the lossless join property.

Copyright © 2011 Ramez Elmasri and Shamkant Navathe Lossless (Non-additive) Join Property of a Decomposition (c) Case 2: Decomposition of EMP_PROJ into EMP, PROJECT, and WORKS_ON satisfies test.

Copyright © 2011 Ramez Elmasri and Shamkant Navathe Testing Binary Decompositions for Lossless Join Property Binary Decomposition: Decomposition of a relation R into two relations. PROPERTY NJB (Nonadditive join test for binary decompositions): A decomposition D = {R1, R2} of R has the lossless join property with respect to a set of functional dependencies F on R if and only if either The f.d. ((R1 ∩ R2)  (R1- R2)) is in F +, or The f.d. ((R1 ∩ R2)  (R2 - R1)) is in F +.

Copyright © 2011 Ramez Elmasri and Shamkant Navathe

Successive Lossless Join Decomposition Claim 2 (Preservation of non-additivity in successive decompositions): If a decomposition D = {R1, R2,..., Rm} of R has the lossless (non-additive) join property with respect to a set of functional dependencies F on R, and if a decomposition Di = {Q1, Q2,..., Qk} of Ri has the lossless (non-additive) join property with respect to the projection of F on Ri, then the decomposition D2 = {R1, R2,..., Ri-1, Q1, Q2,..., Qk, Ri+1,..., Rm} of R has the non-additive join property with respect to F.

Copyright © 2011 Ramez Elmasri and Shamkant Navathe Dependency Preservation Property of a Decomposition Definition: Given a set of dependencies F on R, the projection of F on R i, denoted by p Ri (F) where R i is a subset of R, is the set of dependencies X  Y in F + such that the attributes in X υ Y are all contained in R i. Hence, the projection of F on each relation schema R i in the decomposition D is the set of functional dependencies in F +, the closure of F, such that all their left- and right-hand-side attributes are in R i.

Copyright © 2011 Ramez Elmasri and Shamkant Navathe Dependency Preservation Property of a Decomposition(cont.) Dependency Preservation Property: A decomposition D = {R1, R2,..., Rm} of R is dependency-preserving with respect to F if the union of the projections of F on each Ri in D is equivalent to F; that is ((  R1 (F)) υ... υ (  Rm (F))) + = F + Claim 1: It is always possible to find a dependency- preserving decomposition D with respect to F such that each relation Ri in D is in 3nf.

Copyright © 2011 Ramez Elmasri and Shamkant Navathe Dependency-Preserving Decomposition into 3NF Schemas Algorithm 16.4: Relational Synthesis into 3NF with Dependency Preservation (Relational Synthesis Algorithm) Input: A universal relation R and a set of functional dependencies F on the attributes of R. 1. Find a minimal cover G for F (use Algorithm 16.2); 2. For each left-hand-side X of a functional dependency that appears in G, create a relation schema in D with attributes {X υ {A1} υ {A2}... υ {Ak}}, where X  A1, X  A2,..., X  Ak are the only dependencies in G with X as left-hand-side (X is the key of this relation) ; 3. Place any remaining attributes (that have not been placed in any relation) in a single relation schema to ensure the attribute preservation property. Claim 3: Every relation schema created by Algorithm 16.4 is in 3NF.

Copyright © 2011 Ramez Elmasri and Shamkant Navathe FD1: Emp_ssn → {Esal, Ephone, Dno} FD2: Pno → { Pname, Plocation} FD3: Emp_ssn, Pno → {Esal, Ephone, Dno, Pname, Plocation} Minimal cover G: {Emp_ssn → Esal, Ephone, Dno; Pno → Pname, Plocation} R1 (Emp_ssn, Esal, Ephone, Dno) R2 (Pno, Pname, Plocation) U(Emp_ssn, Pno, Esal, Ephone, Dno, Pname, Plocation)

Copyright © 2011 Ramez Elmasri and Shamkant Navathe Nonadditive Join Decomposition into BCNF Schemas Algorithm 16.5: Relational Decomposition into BCNF with Lossless (non-additive) join property Input: A universal relation R and a set of functional dependencies F on the attributes of R. 1. Set D := {R}; 2. While there is a relation schema Q in D that is not in BCNF do { choose a relation schema Q in D that is not in BCNF; find a functional dependency X  Y in Q that violates BCNF; replace Q in D by two relation schemas (Q - Y) and (X υ Y); }; Assumption: No null values are allowed for the join attributes.

Copyright © 2011 Ramez Elmasri and Shamkant Navathe

Dependency-Preserving and Nonadditive (Lossless) Join Decomposition into 3NF Schemas Algorithm 16.6 Relational Synthesis into 3NF with Dependency Preservation and Lossless (Non-Additive) Join Property Input: A universal relation R and a set of functional dependencies F on the attributes of R. 1. Find a minimal cover G for F (Use Algorithm 16.2). 2. For each left-hand-side X of a functional dependency that appears in G, create a relation schema in D with attributes {X υ {A1} υ {A2}... υ {Ak}}, where X  A1, X  A2,..., X –>Ak are the only dependencies in G with X as left-hand-side (X is the key of this relation). 3. If none of the relation schemas in D contains a key of R, then create one more relation schema in D that contains attributes that form a key of R. (Use Algorithm 16.4a to find the key of R)

Copyright © 2011 Ramez Elmasri and Shamkant Navathe FD1: Emp_ssn → {Esal, Ephone, Dno} FD2: Pno → { Pname, Plocation} FD3: Emp_ssn, Pno → {Esal, Ephone, Dno, Pname, Plocation} Minimal cover G: {Emp_ssn → Esal, Ephone, Dno; Pno → Pname, Plocation} U(Emp_ssn, Pno, Esal, Ephone, Dno, Pname, Plocation) R1 (Emp_ssn, Esal, Ephone, Dno) R2 (Pno, Pname, Plocation) R3 (Emp_ssn, Pno)

Copyright © 2011 Ramez Elmasri and Shamkant Navathe Problems with NULL Values

Copyright © 2011 Ramez Elmasri and Shamkant Navathe Problems with NULL Values

Copyright © 2011 Ramez Elmasri and Shamkant Navathe Problems with Dangling Tuples

Copyright © 2011 Ramez Elmasri and Shamkant Navathe Problems with Dangling Tuples

Copyright © 2011 Ramez Elmasri and Shamkant Navathe Discussion of Normalization Algorithms Problems: The database designer must first specify all the relevant functional dependencies among the database attributes. These algorithms are not deterministic in general. It is not always possible to find a decomposition into relation schemas that preserves dependencies and allows each relation schema in the decomposition to be in BCNF (instead of 3NF as in Algorithm 16.6).

Copyright © 2011 Ramez Elmasri and Shamkant Navathe

Multivalued Dependencies and Fourth Normal Form Revisited Definition: A relation schema R is in 4NF with respect to a set of dependencies F (that includes functional dependencies and multivalued dependencies) if, for every nontrivial multivalued dependency X — >> Y in F +, X is a superkey for R.

Copyright © 2011 Ramez Elmasri and Shamkant Navathe

Multivalued Dependencies and Fourth Normal Form Revisited Lossless (Non-additive) Join Decomposition into 4NF Relations: PROPERTY NJB’ The relation schemas R 1 and R 2 form a lossless (non-additive) join decomposition of R with respect to a set F of functional and multivalued dependencies if and only if (R 1 ∩ R 2 ) — >> (R 1 - R 2 ) or by symmetry, if and only if (R 1 ∩ R 2 ) — >> (R 2 - R 1 )).

Copyright © 2011 Ramez Elmasri and Shamkant Navathe Multivalued Dependencies and Fourth Normal Form Revisited Algorithm 16.7: Relational decomposition into 4NF relations with non-additive join property Input: A universal relation R and a set of functional and multivalued dependencies F. 1.Set D := { R }; 2.While there is a relation schema Q in D that is not in 4NF do { choose a relation schema Q in D that is not in 4NF; find a nontrivial MVD X — >> Y in Q that violates 4NF; replace Q in D by two relation schemas (Q - Y) and (X υ Y); };